Question

If x + y = 1, then the value of x3 + y3 + 3xy is
(1) 1 (2) 0
(3) 2 (4) 3

Answer 1

x+y = 1
(x+y)³ = (1)³
x³ + y³ + 3xy(x+y) = 1
x³ + y³ + 3xy(1) = 1
x³ + y³ + 3xy = 1

(a+b)³ = a³ + b³ +3ab(a+b)

Answer 2

We know that this problem can be easily solved by the help of identities

Given :
$x + y = 1$
Find:
${x}^{3} + {y}^{3} + 3xy$
We also know that for that we had to cube the expression

Take cube both sides of given expression

${(x + y)}^{3} = ( {1)}^{3} \\ \\ open \: identity \\ \\ {x}^{3} + {y}^{3} + 3xy(x + y) = 1 \\ \\ from \: given \: expression \:put \: x + y = 1 \\ \\ \\ {x}^{3} + {y}^{3} + 3xy(1) = 1 \\ \\ {x}^{3} + {y}^{3} + 3xy = 1$
Thus,Option (1) is correct.

Hope it helps you.